I found solution IZHO p:2 The problem doesn't very hard. assume that AB+DE=x, BC+EF=y, CD+FA=z then we have AD^2=xz, BE^2=xy, CF^2=yz. Therefore, AD+BE+CF=<x+y+z=P(hexagon)

Egamberganov wrote: I found solution IZHO p:2 The problem doesn't very hard. assume that AB+DE=x, BC+EF=y, CD+FA=z then we have AD^2=xz, BE^2=xy, CF^2=yz. Therefore, AD+BE+CF=<x+y+z=P(hexagon) The official solution was the same.

Let $BE$ intersect $AD$ at point $M$, $CF$ intersect $AD$ at point $N$, $BE$ intersect $CF$ at point $K$. $S(ABDE)+S(BCEF)+S(CDFA)= \frac{1}{2}hP = \frac{1}{2} BE*AD*sin(NMK)+ \frac{1}{2} BE*CF*sin(MKN) + \frac{1}{2} AD*CF*sin(MNK)(*)$ $h=BE*sin \angle BEF = BE*sin \angle BED$, so $ \angle CBE=\angle BEF=\angle BED=\angle ABE= \beta$, $ \angle BCF=\angle DCF=\angle AFC=\angle CFE= \alpha$, $ \angle CDA=\angle ADE=\angle BAD=\angle DAF= \gamma$, $\alpha +\beta + \gamma =\pi$. Hence $sin(NMK)=sin \alpha = \frac{h}{CF}$, $sin(KNM)=sin \beta = \frac{h}{BE}$, $sin(NKM)=sin \gamma = \frac{h}{AD}$. $AD = x$, $BE = y$, $CF = z$. So $(*) = \frac{1}{2}hP = \frac{1}{2}h(\frac{xz}{y}+ \frac{xy}{z}+ \frac{yz}{x}) \geq \frac{1}{2}h(x+y+z)$, because $(xz)^{2}+ (xy)^{2}+(yz)^{2} \geq xyz(x+y+z)$, which follows from Muirhead's inequality. So $P \geq AD+BE+CF$. $QED.$

Egamberganov wrote: I found solution IZHO p:2 The problem doesn't very hard. assume that AB+DE=x, BC+EF=y, CD+FA=z then we have AD^2=xz, BE^2=xy, CF^2=yz. Therefore, AD+BE+CF=<x+y+z=P(hexagon) from where did you get this? AD^2=xz, BE^2=xy, CF^2=yz.

HAmletAsadullayevCaucasus wrote: Egamberganov wrote: I found solution IZHO p:2 The problem doesn't very hard. assume that AB+DE=x, BC+EF=y, CD+FA=z then we have AD^2=xz, BE^2=xy, CF^2=yz. Therefore, AD+BE+CF=<x+y+z=P(hexagon) from where did you get this? AD^2=xz, BE^2=xy, CF^2=yz. One of possible ways: $AD$ intersects $CF$ at $N$, $AD$ intersects $BE$ at $M$. $S(ABDE)=\frac{1}{2} h(AB+DE) = \frac{1}{2} AD * BE * sin M$. $S(ACDF)=\frac{1}{2} h(CD+AF) = \frac{1}{2} AD * CF * sin N$. So $h^{2}xz=AD^{2}*BE*CF*sin N * sin M = AD^{2} * h^{2}$, because $CF * sin M = BE * sin N = h$. Therefore $AD^{2}=xz$.

how did you find CF * sin M = BE * sin N = h

Denote the common distance by $h$. The given condition implies that $D$ is equidistant from $AB$ and $AF$, i.e. $AD$ bisects $\angle BAF$. SImilarly, $BE$, $CF$, $DA$, $EB$ and $FC$ bisect the corresponding angles in the hexagon $ABCDEF$. Having in mind the parallel lines, we can write $\angle BAD = \angle DAF = \angle ADE = \angle ADC = \alpha$, $\angle ABE = \angle CBE = \angle BEF = \angle BED = \beta$ and hence $\angle BCF = \angle DCF = \angle CFA = \angle CFE = 180^{\circ} - \alpha - \beta$. Now if $AD \cap BE = M$ and $AD \cap CF = N$, then $\angle AMB = \angle AFC = 180^{\circ} - \alpha - \beta$ and $\angle DNC = BED = \beta$. Now we have $\frac{h}{2}(AB+DE) = S_{ABDE} = \frac{\sin(180^{\circ}-\alpha-\beta)}{2}\cdot AD \cdot BE = \frac{h}{2CF} \cdot AD \cdot BE$, hence $AB + DE = \frac{AD \cdot BE}{CF}$. Similarly $BC + EF = \frac{BE \cdot CF}{AD}$ and $CD + AF = \frac{CF \cdot AD}{BE}$. Hence with $x=AD$, $y=BE$, $z=CF$ it is now enough to prove that $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \geq x+y+z$. By clearing the denominators we want $(xy)^2 + (yz)^2 + (zx)^2 \geq (xy)(xz) + (yz)(yx) + (zx)(zy)$ which follows from the well known $a^2 + b^2 + c^2 \geq ab + bc + ca$.